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Rudik [331]
3 years ago
13

Is it possible to form a triangle with side lengths 5, 3, and 5 units?

Mathematics
2 answers:
7nadin3 [17]3 years ago
8 0

Answer:

Yes

Step-by-step explanation:

Isosceles triangle with base of 3 units and sides of 5 units

Artist 52 [7]3 years ago
5 0

Answer:

yes

Step-by-step explanation:

use a+b > c, a+c > b, and b+c > a if two sides are bigger than one side for every combination then you can form a triangle

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At the school festival ,heather can buy 25 game tickets for 10 dollars, or she can pay 0.50 cent per game ticket.whick option ha
irga5000 [103]

Answer:

The 10 dollars for 25 tickets

Step-by-step explanation:

7 0
3 years ago
assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28
melomori [17]

Given:

x and y are both differentiable functions of t.

4x^2+3y^3=28

x=-1\text{ and }\dfrac{dy}{dt}=8

To find:

The value of \dfrac{dx}{dt}.

Solution:

We have,

4x^2+3y^3=28       ...(i)

At x=-1,

4(-1)^2+3y^3=28

4+3y^3=28

3y^3=28-4

3y^3=24

Divide both sides by 3.

y^3=8

Taking cube root on both sides.

y=2

So, y=2 at x=-1.

Differentiate (i) with respect to t.

4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0

Putting x=-1, y=2 and \dfrac{dy}{dt}=8, we get

4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0

-8\dfrac{dx}{dt}+9(4)(8)=0

-8(\dfrac{dx}{dt}-9(4))=0

Divide both sides by -8.

\dfrac{dx}{dt}-36=0

\dfrac{dx}{dt}=36

Therefore, the value of 4x^2+3y^3=28 is 36.

6 0
3 years ago
A pair of fair dice is cast. what is the probabiliy that the sum of the numbers falling uppermost is 9, given that at least one
WITCHER [35]
If a pair of fair dice is cast, the probability that the <span>sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is given by:

\frac{P(sum\, of\, 9\, \cup\,3)}{P(3)}

The number of outcomes of sum of 9 where at last one is 3 is (3, 6) and (6, 3) = 2.

</span>The number of outcomes of last one of the numbers falling uppermost is a 3 is (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3) and (6, 3) = 11.

Therefore, <span>the probabiliy that the sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is 2 / 11.</span>
5 0
4 years ago
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3 years ago
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Ms. Sublett's math class is putting three
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Option 4 is your answer
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